gabriel wrote:samirpandeyit62 wrote:What will be the remainder when 13^7 + 14^7 + 15^7 + 16^7 is divided by 58?
57
1
30
0
28
zero
Lets consider this example before we proceed to the actual problem
7 can be written as a sum of two numbers, lets say 2 and 5 in this case
So,
7/3 is also 2/3 + 5/3 , where
the remainder for 2/3 is 2
the remainder for 5/3 is 2
hence the combined remainder is 2+2 =4 which is the same if 7 is divided by 3.
(note if the combined remainder is bigger than the divisor then we further divide if by the divisor to get the actual remainder).
Similarly in the above case we can expect [58]^7/58
to be written as
(13^7)/58 + (14^7)/58 + (15^7)/58 + (16^7)/58
so its nothing but [58]^7 / 58 hence the remainder is 0.
